The Collision Spectrum

An ordinary mean gives every value the same share. Add eight numbers and divide by eight. Here, the digit function chooses the shares.

In base five, they look like this.

Each bar belongs to one Dirichlet $L$-function value. Its height tells us how much the squared magnitude of that value counts in a weighted mean. The shares come from the boundaries between digits in long division.

The Collision Spectrum and the L-Function Landscape develops the factorization and the exact base-five fourth moment. This preprint also determines the total diagonal weight at every odd prime base. That total makes the averaging rule exact.

To calculate what these weights give us, start with the collision table.

The twenty entries

A collision occurs when multiplication moves a remainder into another remainder that produces the same digit. The Collision Invariant gives a finite table for the deviation of this count from $\lfloor p/b\rfloor$. At lag one, the entry depends only on the last two base-$b$ digits of the prime $p>b^2$.

Here is the table in base five. The row and column labels are the two digits. The entries are ordinary signed integers.

First digit	Ends in 1	Ends in 2	Ends in 3	Ends in 4
0	0	1	0	3
1	0	−1	2	−1
2	0	1	−2	−1
3	0	−3	0	−1
4	−4	−1	−2	−1

The column ending in zero is absent because its residues are divisible by five.

For a concrete entry, take the denominator $29$. Multiplication by five gives eight collisions among its twenty-eight nonzero remainders. Subtract $\lfloor29/5\rfloor=5$ and the deviation is $3$. Since $29=104_5$, it belongs in cell $04$. That is the $3$ at the top right.

Now read down the first column. Four zeros and a $-4$. Its mean is $-4/5$. Subtracting that mean leaves four copies of $4/5$ and one $-16/5$. Their squares add to

$$4\left(\frac45\right)^2+\left(\frac{16}{5}\right)^2=\frac{64}{5}.$$

Do the same in the other three columns. Their means are $-3/5$, $-2/5$ and $-1/5$. The figure shows every centered entry and each column’s sum of squares.

I write the centered entries as $S^\circ(a)$ and call their sum of squares the collision energy. Here it is

$$E_5=\frac{64}{5}+\frac{56}{5}+\frac{56}{5}+\frac{64}{5}=48.$$

The sum includes all twenty residue classes coprime to $25$. Centering is essential. Squaring the original integers would give a different number.

Eight infinite series

A Dirichlet character assigns a complex weight to each residue class, with one rule. The weight of a product is the product of the weights. Modulo $25$, these weights repeat every twenty-five integers and are zero at multiples of five.

Use such a pattern to weight the harmonic series,

$$L(1,\chi)=\chi(1)+\frac{\chi(2)}2+\frac{\chi(3)}3+\cdots.$$

For a nontrivial character, the weights in a complete period sum to zero. The cancellation makes this series converge. Its value is a complex number. The notation $|L(1,\chi)|$ means its magnitude, its distance from zero.

These are the same character weights that appear in Euler products over primes. That connection gives $L$-functions their place in the analytic theory of primes.

The character also reads the finite table. Multiply each centered entry by the conjugate character weight and average over the twenty entries. The result, $\widehat S^\circ(\chi)$, is one Fourier coefficient of the table. The bar in the formulas below denotes complex conjugation.

The Collision Transform explains which coefficients can survive. Reflection removes the even characters. Column centering removes characters whose pattern already repeats modulo $5$. In base five, this leaves eight possible channels. They are the primitive odd characters modulo $25$.

We now have two numbers attached to the same character. One comes from an infinite series. The other comes from twenty entries in a table.

The coefficient as a product

There are two finite sums to look at first.

The generalized Bernoulli number $B_{1,\overline\chi}$ weights each residue $a$ by $a\overline\chi(a)$, adds over the twenty coprime classes and divides by $25$. A classical identity gives its magnitude,

$$|B_{1,\overline\chi}|=\frac5\pi\,|L(1,\chi)|.$$

A finite sum has given us the size of the infinite one.

For the other sum, mark the equal-digit positions $00,11,22,33,44$ in base five. In ordinary notation these are $0,6,12,18,24$. At each position $n$, take the difference $\overline\chi(n+1)-\overline\chi(n)$ and add the five differences. This is the diagonal factor $S_G(\chi)$. It records how the character weights change across those digit boundaries.

Then the coefficients factor.

$$\widehat S^\circ(\chi) =-\frac{B_{1,\overline\chi}\,\overline{S_G(\chi)}}{20}.$$

This is the decomposition theorem at base five. The Bernoulli sum supplies the $L$-value. The diagonal sum supplies the contribution from the digit geometry. Their product gives the Fourier coefficient, including its phase.

The proof follows the floor jumps that define the collision table. Their character sums turn into Bernoulli terms. The part depending only on the final digit cancels, leaving the differences along the equal-digit diagonal. That is where the product comes from.

A weighted mean for every odd prime base

The factorization holds for every odd prime base $b$. The denominator $20$ becomes $b(b-1)$, the number of coprime classes modulo $b^2$.

Parseval’s identity now lets us add the squares. It says that the total squared size of a finite signal agrees with the total squared size of its Fourier coefficients, with the normalization accounted for. Here each coefficient has an $L$-value inside it. Substituting the factorization gives

$$\sum_{\chi\in\mathcal P_b}|L(1,\chi)S_G(\chi)|^2 =\pi^2\frac{b-1}{b}E_b.$$

The set $\mathcal P_b$ consists of the primitive odd characters modulo $b^2$. The energy $E_b$ sums $|S^\circ(a)|^2$ over all $b(b-1)$ coprime classes.

So the table evaluates a weighted second moment of the $L$-values. Each squared $L$-value receives a weight supplied by the digit diagonal. The total of these weights is exact too,

$$\sum_{\chi\in\mathcal P_b}|S_G(\chi)|^2=2b(b-1)^2.$$

Divide by that total and the weights add to one. This gives the shares in the opening figure. At base five the unnormalized total is $160$, and the resulting weighted mean of the squared $L$-value magnitudes is $6\pi^2/25$.

The exact total follows from character orthogonality. When the squared diagonal sums are expanded, we compare every pair of positions on the diagonal. Matching positions contribute. The two remaining congruence contributions cancel. Counting the surviving pairs gives $2b(b-1)^2$ at every odd prime base.

We therefore know both the weights and the weighted mean. The digit function supplies the rule for averaging, and the collision energy evaluates the answer.

How the diagonal weights and the $L$-values vary together as the prime base grows remains open. At base five, their relationship can be settled in every channel.

The second copy of the L-value

At base five, the two finite factors have proportional magnitudes,

$$|S_G(\chi)|=\sqrt5\,|B_{1,\overline\chi}|.$$

The proof is a polynomial identity. The residue $2$ generates all twenty coprime classes modulo $25$, so a character is determined by its value at $2$. For each of our eight characters, that value is a primitive twentieth root of unity. Both finite sums become polynomials in the same root. Reducing those polynomials gives the equality above.

The Bernoulli factor already carries $|L(1,\chi)|$. Now the diagonal factor carries a second copy. Multiplying them gives

$$|\widehat S^\circ(\chi)| =\frac{5\sqrt5}{4\pi^2}\,|L(1,\chi)|^2.$$

Each Fourier magnitude is a squared $L$-value with the same scaling constant. Squaring those magnitudes once more produces fourth powers.

Call their sum $M_5$,

$$M_5=\sum_{\chi\in\mathcal P_5}|L(1,\chi)|^4.$$

Both members of each conjugate pair are counted. There are eight terms. Parseval says their corresponding squared Fourier magnitudes add to $E_5/20$. Substituting the coefficient formula and rearranging gives $M_5=(4\pi^4/625)E_5$.

The energy is the $48$ we calculated from the table. Hence

$$M_5=\frac{4\pi^4}{625}\,48=\frac{192\pi^4}{625}.$$

I began with a question about two remainders producing the same digit. That question gave a finite table. Its symmetry gave a character expansion. The factorization now tells us what those coefficients contain.

Here, at base five, we can carry the calculation to an exact answer. Twenty entries, four column means and a sum of squares have evaluated the fourth moment of eight infinite series. The number that made it possible was $48$.